Exercise 2_2_1: Across the Tidyverse

library(tidyverse)
library(sjlabelled)

gp_covid <- 
  read_csv2("./data/ZA5667_v1-1-0.csv") %>%
  set_na(na = c(-99, -77, -33, 98))

1

Use across() to dichotomize the trust variables hzcy044a:hzcy052a. The value 1 should remain, all others should be 0.

Clues

You have to wrap it into the mutate() function and then use recode().

solution

gp_covid <- 
  gp_covid %>% 
  mutate(
    across(
      hzcy044a:hzcy052a,
      ~recode(
        .x,
        `5` = 0,
        `4` = 0,
        `2` = 0,
        `1` = 1
      )
    )
  )

2

Please check if all went as expected.

Clues

There are several ways of doing that.

1. Either print tables like on the slides using the table() function.
2. Or use select() to select only the manipulated columns and print the whole data frame using View(), glimpse() or the plain console output.

solution

table(gp_covid$hzcy044a)

## 
##    0    1    3 
## 2693   43  329

table(gp_covid$hzcy047a)

## 
##    0    1    3 
## 2878   29  188

table(gp_covid$hzcy052a)

## 
##    0    1    3 
## 2779   25  303

gp_covid %>% 
  select(hzcy044a:hzcy052a) %>% 
  glimpse()

## Rows: 3,765
## Columns: 9
## $ hzcy044a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 3, 0, NA, 0, 3, NA, NA, 0, NA, 0, NA, 0, 0, NA, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0, 0, 0, 0, 3,~
## $ hzcy045a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 3, NA, 3, 3, NA, NA, 0, NA, 0, NA, 0, 0, NA, 0, NA, NA, 3, 0, NA, NA, 0, NA, 0, NA, 0, NA~
## $ hzcy046a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 3, 0, 0, NA, 3, 3, NA, NA, 0, NA, 0, NA, 3, 0, NA, 0, 0, 3, 3, 0, NA, NA, 3, NA, 0, 0, 3, 3, 3,~
## $ hzcy047a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 3, NA, 0, 0, 0, 0, 3, ~
## $ hzcy048a <dbl> NA, 0, 0, 0, NA, 0, 3, 0, NA, 3, 0, 1, NA, 0, 0, NA, NA, 0, NA, 0, NA, 3, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 3, 3, 0, 0, 3, ~
## $ hzcy049a <dbl> NA, 0, 3, 0, NA, 0, 3, 0, NA, 0, 1, 0, NA, 0, 0, NA, NA, 0, NA, NA, NA, 3, 0, 0, 0, 0, 0, 0, 3, NA, NA, 0, NA, 1, 3, 0, 0, 3,~
## $ hzcy050a <dbl> NA, 0, 0, 0, NA, 0, 3, 0, NA, 0, 0, 1, NA, 0, 0, NA, NA, 0, NA, 0, NA, 3, 0, 0, 0, 0, 0, 0, 0, NA, NA, 0, NA, 0, 3, 0, 0, 3, ~
## $ hzcy051a <dbl> NA, 0, 0, 0, NA, 3, 0, 0, NA, 0, 0, 3, NA, 0, 0, NA, NA, 0, NA, 3, NA, 0, 0, 0, 0, 0, 0, 3, 0, NA, NA, 3, NA, 0, 3, 0, 0, 3, ~
## $ hzcy052a <dbl> NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, 0, NA, 0, 0, NA, NA, 0, NA, 0, NA, 0, 0, 0, 0, 0, 3, 3, 0, NA, NA, 0, NA, 0, 0, 0, 0, 0, ~

3

Use the median() function to calculate an aggregated variable of all trust variables for all observations in the data.

Clues

For this task you need to first use rowwise() on your data and calculate the values, amongst others, applying the c_across() function. Oh, and don’t forget to finally use ungroup.

solution

gp_covid <- 
  gp_covid %>% 
  rowwise() %>% #<<
  mutate(
    median_trust = 
      median(
        c_across(hzcy044a:hzcy052a),
        na.rm = TRUE
      )
  ) %>% 
  ungroup()

table(gp_covid$median_trust)

## 
##    0  0.5    1  1.5    2    3 
## 2795    9  102   17    2  235